The general term of ${(2a-3b)}^{-\frac{1}{2}}$ is

$\begin{array}{l}{(2a-3b)}^{-\frac{1}{2}}\\ =\frac{n(n+1)(n+2)(n+3)\cdots (n+r-1)}{r!}{x}^r\\ {(2a-3b)}^{\frac{-1}{2}}={\left(2a\right)}^{\frac{-1}{2}}{\left(1-\frac{3b}{2a}\right)}^{-\frac{1}{2}}\\ \Rightarrow \frac{1}{\sqrt{2}a}(1+\frac{1}{2}\left(\frac{3b}{2a}\right)+\frac{\frac{1}{2}\times \frac{3}{2}}{2!}{\left(\frac{3b}{2a}\right)}^2+\cdots )\end{array}$

General term $\left[\frac{\left(\frac{1}{2}\left(\frac{1}{2}+1\right)(\frac{1}{2}+2)\cdots \left(\frac{1}{2}+r-1\right)\right)}{r!}{\left(\frac{3b}{2a}\right)}^r\right]\frac{1}{\sqrt{2a}}$

$=\frac{\frac{1}{2}\times \frac{3}{2}\times \frac{5}{2}\cdots \left(\frac{2r-1}{2}\right){\left(\frac{3b}{2a}\right)}^r}{r!}.\frac{1}{\sqrt{2a}}$

$=\left(\frac{{\displaystyle \mathrm{1.3.5}\cdots (2r-1)}}{{\displaystyle 2^r.r!}}{\left(\frac{{\displaystyle 3b}}{{\displaystyle 2a}}\right)}^r\right).\frac{1}{\sqrt{2a}}$

$=\frac{\mathrm{1.3.5}\cdots (2r-1)}{2^r.r!}{\left(\frac{{\displaystyle 3b}}{{\displaystyle 2a}}\right)}^r.\frac{1}{\sqrt{2a}}$